Pneumatics Training System, Model 6081

Fluid Power

Pneumatics Fundamentals

Courseware Sample 31290-F0

Order no.: 31290-00 First Edition Revision level: 01/2015

By the staff of Festo Didactic

© Festo Didactic Ltée/Ltd, Quebec, Canada 1997 Internet: www.festo-didactic.com e-mail: [email protected]

Printed in Canada All rights reserved ISBN 978-2-89289-383-0 (Printed version) ISBN 978-2-89640-652-4 (CD-ROM) Legal Deposit – Bibliothèque et Archives nationales du Québec, 1997 Legal Deposit – Library and Archives Canada, 1997

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The purchaser shall be entitled to use the work to train his/her staff at the purchaser's site/location and shall also be entitled to use parts of the copyright material as the basis for the production of his/her own training documentation for the training of his/her staff at the purchaser's site/location with acknowledgement of source and to make copies for this purpose. In the case of schools/technical colleges, training centers, and universities, the right of use shall also include use by school and college students and trainees at the purchaser's site/location for teaching purposes.

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Safety and Common Symbols

Symbol Description

Protective conductor terminal

Frame or chassis terminal

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On (supply)

Off (supply)

Equipment protected throughout by double insulation or reinforced insulation

In position of a bi-stable push control

Out position of a bi-stable push control

We invite readers of this manual to send us their tips, feedback and suggestions for improving the book.

Please send these to [email protected].

The authors and Festo Didactic look forward to your comments.

III

Table of ContentsIntroduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V

Courseware Outline

Pneumatics Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . VII

Electrical Control of Pneumatic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . XI

Pneumatics Applications – PLC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . XV

Sample Exercise from Pneumatics Fundamentals

Ex. 4-1 Indirect Control Using Pilot-Operated Valve . . . . . . . . . . . . . . . . . . . 3

To introduce the operation of pilot-operated directional control valves.To learn about construction and classification. To show the advantagesof indirect control in demo circuits using a long line device, cylindersand a 4-way, 5-port, 2-position pilot-operated directional control valve.

Sample Exercise from Electrical Control of Pneumatic Systems

Ex. 3-1 Basic Memory and Priority Electropneumatic Circuits . . . . . . . . . . 15

To show how a directional valve can memorize a signal and maintaina position. To demonstrate how to pneumatically lock and unlock anelectropneumatic circuit. Comparison between air-locked andelectrically-locked circuits. Introduction to limit switches.

Sample Exercise from Pneumatics Applications – PLC

Ex. 6 Counting of Pneumatic Actuator Cycles . . . . . . . . . . . . . . . . . . . . . 31

Connection and operation of a PLC-controlled pneumatic system thatmakes a motor rotate 200 turns and then reciprocates a cylinder5 times.

Sample Exercise from Servo/Proportional Control of Pneumatic Systems

Ex. 5 Closed-Loop Position Control, Proportional-Plus-Integral Mode . . 43

Description of the integral control mode. Definition of the terms integralgain, overshoot, and oscillation. Advantages and disadvantages ofintegral control. Comparison of the proportional, integral andproportional-plus-integral control modes.

Other samples extracted from Pneumatics Fundamentals

Unit Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Instructor’s Guide Sample Extract from Electric Control of PneumaticsSystems

Unit 4 Industrial Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Bibliography

IV

V

Introduction

The Lab-Volt Pneumatics Training System, Model 6081, is a modular program inpneumatics and its applications. The system is divided into five subsystems:Pneumatics Fundamentals, Electrical Control of Pneumatic Systems, PneumaticsApplications – PLC, Servo/Proportional Control of Pneumatic Systems.

In Pneumatics Fundamentals, the students are introduced to the basic principlesand components of pneumatics. Electrical Control of Pneumatic Systems coverselectrical control of pneumatic systems with ladder diagrams. PneumaticsApplications – PLC expands upon the others with pneumatics applicationsdemonstrating programmable logic controllers (PLCs). In Servo/ProportionalControl of Pneumatic Systems, students are introduced to servo-proportionalcontrol systems and their associated circuitry.

Most of the components used for electrical control of the Lab-Volt PneumaticsTraining System are also intended to be used with the Lab-Volt Hydraulics TrainingSystem, Model 6080, allowing interconnection of both systems to perform morecomplete functions.

Innovative design

Engineered for extreme ease of use, the basic system comes with a work surfaceassembly consisting of a solid metal, universal drip-tray hinged to a perforated,tiltable work surface on which pneumatic components can be mounted. The worksurface can be configured to accommodate a wide variety of space and teachingneeds. Lying flat over the drip tray, or tilted at 45 , the work surface provides alarge area on which pneumatic components and space-expanding work surfacescan be mounted at any location, either at 0 or 90 .

Mounting and removal of components is especially easy with push-lock fastenersthat snap effortlessly into the perforations of the work surface.

Although the system is designed to operate atop a regular work table, an optionalbench is available to provide mobility and storage space. Mounted on four heavy-duty, swivelling, lockable castors, the bench provides shelving for extra worksurfaces and components. Optional dressing panels are also available to fullyenclose the bench and provide storage area for components. The work surfacesmay also be linked together for more complete electro-pneumatic/hydraulic circuits.By utilizing the optional work surfaces, exercise setups can be saved. Allcomponents meet industrial safety standards, are identified with the proper ANSIsymbols on their base, and are equipped with quick-connect fittings.

Virtual Laboratory Equipment

All components contained in Pneumatics Fundamentals and Electrical Control ofPneumatic Systems can be simulated by the Windows™-based simulation softwareLab-Volt Pneumatics Simulation Software (LVSIM®-PNEU), Model 6485.

VI

Courseware

The Pneumatics courseware consists of a student manual for each subsystem andan instructor guide. Student manuals are divided into several units, each consistingof a series of hands-on exercises dealing with one of the pneumatics fields. Eachexercise provides a clearly stated objective, a discussion, textbook references, asummary of the exercise procedure, a conclusion, and a set of review questions.A ten-question test at the end of each unit allows the student to verify what waslearned in the unit.

The instructor guides contain the measurement results as well as the answers foreach hands-on exercise of the student manuals, and the answers to the unit testquestions.

The Pneumatics courseware also includes a reference textbook, optional videotapes and courseware figures. These figures are available in the form oftransparencies and on CD-ROM.

PNEUMATICS FUNDAMENTALS

Courseware Outline

VII

Unit 1 Introduction to Pneumatics

Explore the trainer and components included in the Lab-Volt PneumaticsTraining System. Safety rules, component identification, description andgeneral operation. Introduction to air conditioning and distributingequipment.

Ex. 1-1 Familiarization with the Lab-Volt Pneumatics Trainer

Description of the Lab-Volt Pneumatics Trainer. Configuration ofthe work surface. Identification of the various components.Familiarization with the symbols, characteristics and uses of eachcomponent. Safety rules.

Ex. 1-2 Introduction to Pneumatics

To introduce pneumatic power characteristics, applications,advantages and disadvantages. To investigate a demo circuitusing a directional control valve and a cylinder.

Ex. 1-3 Air Conditioning and Distributing Equipment

To introduce the Conditioning Unit and its components: shutoffvalves, filter, pressure gauge, pressure regulator and muffler. Tolearn about receivers, accumulators and safety relief valves. Toobserve the effect of friction in a demo circuit using anaccumulator, a directional control valve, a flow control valve and acylinder.

Unit 2 Basic Physical Concepts

To introduce pressure, force, volume and flow relationships. Vacuumgeneration. Measurements using pneumatic components. Introduction toflowmeters, needle valves, check valves, flow control valves and vacuumgenerators.

Ex. 2-1 Pressure vs Force Relationship

To introduce the relationship between pressure and force. To verifythe formula F = P x A. To measure the force delivered by a cylinderin demo circuits using a cylinder, a pressure gauge and a loaddevice. To observe that the force exerted on a given surface isdirectly proportional to the pressure applied on this surface.


Courseware Outline

VIII

Ex. 2-2 Pressure vs Volume Relationship

To introduce the relationship between pressure and volume. Toverify the formula (P1×V1)/T1 = (P2×V2)/T2 by compressing air ina cylinder chamber in demo circuits using a cylinder and apressure gauge.

Ex. 2-3 Pressure Drop vs Flow Relationship

To introduce the relationship between pressure drop andgenerated flow. To see the effect of a load on the flow in democircuits using a flowmeter, a flow control valve and a pressuregauge. To introduce flowmeters, needle valves, check valves andflow control valves.

Ex. 2-4 Vacuum Generation

To introduce vacuum generation in demo circuits using avacuum generator, cylinders, an air bearing and a pressuregauge. To demonstrate manometer operating principles bymeasuring the height of a column of water in a demo circuit.

Unit 3 Basic Controls of Cylinders

To introduce components used in fundamental circuits featuring directionalcontrol valves and cylinders. Introduction to the methods for controllingspeed, force and synchronization.

Ex. 3-1 Directional Control Valves

To introduce the operation of directional control valves. To learnabout symbols, operators, construction and classification. To learnabout normally passing and normally non-passing valves. To learnhow to select circuit branches and power sources in demo circuitsusing a flowmeter, a flow control valve and a 3-way, 2-positiondirectional control valve.

Ex. 3-2 Directional and Speed Control of Cylinders

To introduce the operation of cylinders. To learn about symbols,dimension parameters, construction and classification. To learnhow to control the speed of cylinders using flow control valves. Toverify the meter-in and meter-out methods of control in democircuits using flow control valves, directional control and cylinders.


Courseware Outline

IX

Ex. 3-3 Cylinders in Series

To describe the operation of a series circuit and cylindersynchronization. To demonstrate pressure intensification in democircuits using a directional control valve and cylinders.

Ex. 3-4 Cylinders in Parallel

To describe the operation of a parallel circuit, to learn about theextension sequence of parallel cylinders having different loads. Toshow how to synchronize the extension of parallel cylinders indemo circuits using directional control valves, cylinders and flowcontrol valves.

Unit 4 Basic Controls of Pneumatic Motors

To introduce pilot-operated directional control valves and pneumaticmotors. To learn the methods for controlling torque, speed and direction ofrotation of pneumatic motors.

Ex. 4-1 Indirect Control Using Pilot-Operated Valves

To introduce the operation of pilot-operated directional controlvalves. To learn about construction and classification. To show theadvantages of indirect control in demo circuits using a long linedevice, cylinders and a 4-way, 5-port, 2-position pilot-operateddirectional control valve.

Ex. 4-2 Pneumatic Motor Circuits

To introduce symbols, construction and classification of pneumaticmotors. To show how to control torque, direction and speed controlof a motor in a test circuit using directional control valves, flowcontrol valves and a pneumatic motor.

Ex. 4-3 Pneumatic Motor Performance

To introduce how to use manufacturer's data sheets. To learn howto evaluate the performance of a pneumatic motor in demo circuitsusing a flowmeter, a flow control valve and a motor.


Courseware Outline

X

Appendices A Equipment Utilization ChartB Care of the Pneumatics TrainerC Hydraulics and Pneumatics Graphic Symbols D Conversion FactorsE New Terms and Words

Bibliography

We Value Your Opinion!

ELECTRICAL CONTROL OF PNEUMATIC SYSTEMS

Courseware Outline

XI

Unit 1 Introduction to Electrical Control of Pneumatic Systems

An introduction to electrically-controlled pneumatic systems. Description ofthe function of each part of an electrical control circuit.

Ex. 1-1 Familiarization with the Equipment

Identification of the components used for electrical control of theLab-Volt Pneumatics Trainer. Classifying the components as inputelement, controller element, or actuating mechanism.

Unit 2 Electrical Concepts

Basic concepts of electricity. How to read, draw and connect simple ladderdiagrams. Experiment typical basic circuits involving logic function valves.

Ex. 2-1 Basic Electricity

Measurement of the voltage, resistance, and current in anelectrical control circuit. Connection and operation of an electricalcontrol circuit.

Ex. 2-2 Ladder Diagrams

Definition of a ladder diagram. Description of how a ladderdiagram operates and how it relates to the pneumatic equipment.Rules for drawing ladder diagrams. Connection and operation ofbasic ladder diagrams using series (AND) logic, parallel (OR)logic and control relays.

Ex. 2-3 Basic Electrically-Controlled Pneumatic Circuits

To show the advantage of indirect control where a maindirectional valve is actuated by a pressure signal delivered byanother directional valve or by an electrical signal provided by aninput device. How to improve reciprocating time response usinga quick exhaust valve. Introduction to magnetic proximity switchesand solenoid-operated directional valves.

Ex. 2-4 Basic AND and OR Logic Function Circuits

To introduce the AND function valve and the shuttle valve (OR).To assemble and test circuits using these logic functions.


Courseware Outline

XII

Unit 3 Functional Systems

Connection and operation of functional electrically-controlled pneumaticsystems.

Ex. 3-1 Basic Memory and Priority Electropneumatic Circuits

To show how a directional valve can memorize a signal andmaintain a position. To demonstrate how to pneumatically lockand unlock an electropneumatic circuit. Comparison between air-locked and electrically-locked circuits. Introduction to limitswitches.

Ex. 3-2 Multi-Pressure Systems

To use a pressure regulating valve to show how to get a lowerpressure at one point of a circuit while the working pressure of thecircuit remains at a higher value. See how to manage multiple-pressure control in order to create a shift in the force exerted byan actuator in choosing a different pressure setting. Introductionto pressure-switches.

Ex. 3-3 Sequencing Pneumatic Circuits

To learn basic circuits involving sequencing to control actuatorsin a specific order. How to electrically create a sequence withouthaving a sequence valve in the circuit. Introduction to cascadecircuits.

Ex. 3-4 Time-Delay Electropneumatic Applications

To create an alternating circuit to simulate a cycle-operatingapplication featuring a time-delay relay. Learn how to use aircompression to control a time-delay application. Introductionto time-delay relay.

Unit 4 Industrial Applications

To introduce industrial-type circuits and sensors used in differentapplications. Simulate conditions involved and show advantage andflexibility of an electropneumatic control.

Ex. 4-1 Pneumatic Actuator Deceleration Circuits

To reproduce a typical industrial application involving shift invelocity or braking of an actuator. Comparison between air-actuated circuit and electrically-actuated circuit.


Courseware Outline

XIII

Ex. 4-2 Counting of Actuator Cycles

To create an alternating circuit to simulate a cycle-operatingapplication. Learn basic rules involved in that type of control.Introduction to counters.

Ex. 4-3 Industrial Drilling System and Safety Circuits

To build a drilling machine circuit to reproduce typical industrialapplications.

Ex. 4-4 Garbage Compactor Simulation Circuit

To build a garbage compactor circuit to simulate a well-knownapplication as a synthesis of the notions previously learned andhow to set up multiple control devices to make them work properlyin a large electropneumatic circuit.

Unit 5 Troubleshooting

To use simple and logical methods to perform troubleshooting applied toboth electric and pneumatic circuits.

Ex. 5-1 Troubleshooting Electrical Control Circuits

Description of the voltmeter and ohmmeter methods oftroubleshooting an electrical control circuit. Location of instructor-inserted faults in the electrical section of an electrically-controlledsystem.

Ex. 5-2 Troubleshooting Electrically-Controlled Pneumatic Systems

Learning an efficient troubleshooting method for locating faults inan electrically-controlled pneumatic system. Location of instructor-inserted faults in the pneumatic and electrical sections of a circuit.


Courseware Outline

XIV

Appendices A Equipment Utilization ChartB Care of the Pneumatics TrainerC Hydraulics and Pneumatics Graphic Symbols D Ladder Diagram Graphic SymbolsE Conversion FactorsF Trainer Status Verification ProcedureG Time-Delay Relay/Counter SpecificationsH New Terms and Words

Bibliography


PNEUMATICS APPLICATIONS – PLC

Courseware Outline

XV

Exercise 1 Programmable Logic Controller Review

Revision of the PLC relay-type instructions. Entering and testing aprogram that uses relay-type instructions to control the turning onand turning off of two lamps.

Exercise 2 Timer Instructions

Revision of the PLC timer instructions. Entering and testing aprogram that uses timer-on instructions to turn on three lamps in aprogrammed order and for a definite period of time.

Exercise 3 Counter Instructions

Revision of the PLC counter instructions. Entering and testing aprogram that uses two counters in cascade to turn on a lamp afteranother lamp has turned on a definite number of times.

Exercise 4 Latching and Comparison Instructions

Revision of the PLC latching and comparison instructions. Enteringand testing a program that uses latching and counter-drivencomparison instructions to turn on a lamp after another lamp hasblinked a definite number of times.

Exercise 5 Time-Delay Control of Pneumatic Actuators

Connection and operation of a PLC-controlled pneumatic system thatcontinuously reciprocates a cylinder and makes it dwell in twopredetermined positions for some period of time.

Exercise 6 Counting of Pneumatic Actuator Cycles

Connection and operation of a PLC-controlled pneumatic system thatmakes a motor rotate 200 turns and then reciprocates a cylinder5 times.

Exercise 7 Safety Control of Pneumatic Actuators

Connection and operation of a PLC-controlled pneumatic system thatuses a STOP/RESET pushbutton, a pressure switch, and an alarmlamp to provide safety control of a press cylinder.

Exercise 8 PLC-Controlled Clamp and Work System

Connection and operation of an industrial-type clamp and worksystem that monitors the pressure applied by the clamp cylinder toensure the workpiece remains firmly clamped while being worked on.

PNEUMATICS APPLICATIONS – PLC

Courseware Outline

XVI

Exercise 9 Troubleshooting

Location of instructor-inserted faults in the pneumatic and PLC-control sections of the clamp and work system studied in Exercise 8.

Exercise 10 Designing a PLC-Controlled Stamping Machine

Designing a PLC-controlled stamping machine that embosses thinmetal sheets.

Exercise 11 Designing a PLC-Controlled Conveyor System

Designing a PLC-controlled conveyor system that circulatesmanufactured parts and loads them on a packing machine.

Exercise 12 Designing a PLC-Controlled Injection Molding Machine

Designing a PLC-controlled injection molding machine used toproduce plastic dice.

Appendices A Equipment Utilization ChartB Care of the Pneumatics TrainerC Hydraulics and Pneumatics Graphic SymbolsD Ladder Diagram Graphic SymbolsE Conversion FactorsF Trainer Status Verification ProcedureG Troubleshooting ProceduresH New Terms and Words

Bibliography

We Value Your Opinion

SERVO/PROPORTIONAL CONTROL OF PNEUMATIC SYSTEMS

Courseware Outline

XVII

Exercise 1 Introduction to Servo Control Valves . . . . . . . . . . . . . . . . . . . 1-1

Description and operation of servo control valves of the pressure type.Pressure versus voltage characteristic of the trainer pa6429.Introduction to the SETPOINTS Section of the trainer PID Controller.

Exercise 2 Acceleration and Deceleration Control . . . . . . . . . . . . . . . . . . 2-1

Elimination of abrupt starting and stopping of an actuator withacceleration and deceleration control. Introduction to the RAMPGENERATOR of the Trainer PID Controller

Exercise 3 Open-Loop Position Control . . . . . . . . . . . . . . . . . . . . . . . . . . 3-1

Description of open-loop position control systems. Definition of theterms open-loop, closed-loop, and disturbances. Sensing the positionof a cylinder rod. Open-loop control of the trainer cylinder rod position.

Exercise 4 Closed-Loop Position Control, Proportional Mode . . . . . . . . 4-1

Description of closed-loop position control systems. Definition of theterms proportional gain, proportional band, residual error, and manualreset. Advantages and disadvantages of the proportional controlmode.

Exercise 5 Closed-Loop Position Control, Proportional-Plus-Integral Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5-1

Description of the integral control mode. Definition of the termsintegral gain, overshoot, and oscillation. Advantages anddisadvantages of integral control. Comparison of the proportional,integral and proportional-plus-integral control modes.

Exercise 6 Open-Loop Speed Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6-1

Description of open-loop speed control systems. Sensing the speedof a pneumatic motor. Open-loop control of the trainer motor speed,and span setting.

SERVO/PROPORTIONAL CONTROL OF PNEUMATIC SYSTEMS

Courseware Outline

XVIII

Exercise 7 Closed-Loop Speed Control, Proportional-Plus-Integral-Plus-Derivative Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7-1

Description of the derivative control mode. Definition of the termsderivative time, ideal and parallel configurations. Advantages anddisadvantages of derivative control. Description of the proportional-plus-integral-plus-derivative control mode. Comparison of theproportional, proportional-plus-integral, and proportional-plus-integral-plus-derivative control modes.

Exercise 8 Closed-Loop Pressure Control, Proportional-Plus-Integral Mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8-1

Description of open-loop and closed-loop control of circuit pressure.Sensing the pressure using a pressure transducer.

Appendices A Equipment Utilization Chart . . . . . . . . . . . . . . . . . . . . . . . A-1B Trainer Status Verification Procedure . . . . . . . . . . . . . . B-1C Hydraulics and Pneumatics Graphic Symbols . . . . . . . C-1D Pressure Transducer Setting . . . . . . . . . . . . . . . . . . . . . . D-1E Conversion Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E-1F Care of the Pneumatics Trainer . . . . . . . . . . . . . . . . . . . . F-1G New Terms and Words . . . . . . . . . . . . . . . . . . . . . . . . . . . G-1

Bibliography


Sample Exercise

from


3

Exercise 4-1

Indirect Control Using Pilot-Operated Valves

EXERCISE OBJECTIVE

To learn about pilot-operated valves;To show the advantages of indirect control of a single acting cylinder.

DISCUSSION

The main difference between pilot-operated directional control valves and direct-operated valves is in how their spools are shifted. On a pilot-operated valve, an airsignal replaces the mechanical force used to shift the spool in a direct-operatedvalve. Other than this, the housings and spools of both valve types are so similarthat these parts are often interchangeable.

The greatest advantage of a pilot-operated valve is that it permits theremote-actuation of large valves with inexpensive pilot lines. The more expensiveworking lines of the larger valves can then be kept short to save money. Cheaperpilot-lines can be run for some distance without any loss of circuit performance.Since pilot-operated valves do need not be manually actuated, they can becontrolled with outside devices or systems. This makes process automationpossible. Also, because pilots require minimum pressures and volumes to shiftagainst the working pressures, they reduce the delays caused by compressibilityof air and friction in long tubing lines.

Pilot-operated valves can be 3-way and 4-way (4 ports or 5 ports). They can beeither 2-position or 3-position. Usually, 3-way pilot valves are used to remotelycontrol linear or rotary actuators in one direction and then exhaust their workinglines. The 4-way pilot valves are used to remotely control linear and rotaryactuators in two directions, as well as exhausting their working lines.

Pilot-operated valves may move their spools using one or two pilots along with areturn spring. If a 2-position valve uses only one pilot, the pilot moves the valvespool against a spring and into the housing opposite the pilot. The spring will returnthe spool when pilot pressure is removed.

Double-piloted valves have a pilot at each end of the housing. Opposing pilots areused to shift the spool back and forth, but the circuit must exhaust one pilot beforethe other pilot can shift the spool. The lack of return springs in double-pilotedvalves, allows the spool position to be maintained, or memorized, withoutmaintaining pilot pressure.

The pilot-operated valve supplied with your Trainer is a double-piloted, 4-way,5-port, 2-position directional control valve. It is called a 4-way valve instead of a


4

PILOTSYMBOL

PRESSUREPORT

PORTSPILOT

EXHAUSTPORT SYMBOL

15

4

3

2

EXHAUST PORTTO ATMOSPHERE

PRESSURE PORT

24

5 31

PORT BPILOT PILOT

PORT AFROM BRANCH

CIRCUIT CIRCUITTO BRANCH

SYMBOL

5-way valve because one of the exhaust ports is usually not used in a given valveposition. It is illustrated in Figure 4-1.

Figure 4-1. Double-pilot, 4-way, 2-position Directional Control Valve.

The operation of a double-piloted, 4-way, 2-position directional control valve isillustrated in Figure 4-2. When pilot port A is pressurized, ports 1 and 2 areinterconnected through the valve, supplying the branch circuit with compressed air.Ports 3 and 4 are also interconnected through the valve, connecting the branchcircuit to atmosphere. When the spool is shifted by pressurizing pilot port B, ports 1and 4 are interconnected to supply the branch circuit with compressed air.Compressed air is exhausted from the branch circuit to atmosphere throughinterconnected ports 2 and 3. Pilot-operated valves can have manual overrides tomove the spool without pilot pressure for system setup and troubleshooting.

Figure 4-2. Operation of a Double-pilot, 4-way, 2-position Directional Control Valve.

The circuit shown in Figure 4-3 shows that a 3-way, 2-position pilot-operated valveallows the use of shorter expensive working lines. The cheaper pilot-lines can be


5

LONG DELAY

WORKINGPRESSURE

SHORT DELAY

PRESSUREPILOT

PRESSUREWORKING

run for some distance without any loss of circuit performance while minimizingdelays.

Figure 4-3. Indirect Control Using a 3-way, 2-position Piloted Directional Control Valve.

REFERENCE MATERIAL

For additional information on directional control valves, refer to the chapter entitledDirectional Control Valves in the Parker-Hannifin manual Industrial PneumaticTechnology.

Procedure summary

In this exercise, you will verify the operation of a 4-way, 5-port, 2-positiondirectional control valve by operating a double-acting cylinder.

In the second part you will use the long line device to verify that pilot lines requireminimum pressure and volume to shift the spool.

You will verify that priority can be maintained on a position when the pilot portremains pressurized. You will also verify that indirect control reduces delays causedby compressibility of air and friction in long tubing lines.


6

DCV 1

DCV 2

EQUIPMENT REQUIRED

Refer to the Equipment Utilization Chart, in Appendix A of this manual, to obtain thelist of equipment required to perform this exercise.

PROCEDURE

1. Verify the status of the trainer according to the procedure given inExercise 1-2.

2. Retract the piston rod of the Double-Acting Cylinder and connect the circuitshown in Figure 4-4.

Figure 4-4. Schematic Diagram of a Pilot-Operated Circuit.

3. From the schematic diagram shown in Figure 4-4, predict which of thevalves DCV1 or DCV2 controls the extension and/or the retraction of thepiston rod?

4. Open the main shutoff valve and the branch shutoff valves at the manifoldand set the pressure regulator at 100 kPa (or 15 psi) on the regulatedPressure Gauge.


7

5. Push down the button on the directional control valve DCV2 to set thespool in the pilot-operated valve as illustrated in Figure 4-4. The piston rodshould be retracted.

6. Actuate the cylinder using the directional control valves DCV1 and DCV2.Do the valves DCV1 and DCV2 control the extension and retraction of thecylinder piston rod as predicted? If not, explain why.

7. Push down the button on the directional control valve DCV2 and maintainthis position. With your other hand, push down the button on the directionalcontrol valve DCV1. Does the piston rod extend?

Yes No

8. Release the button on the directional control valve DCV2, then push downthe button on the directional control valve DCV1 and maintain this position.With your other hand, push down the button on the directional control valveDCV2. Does the piston rod retract? Explain why.

9. Close the shutoff valves and turn the regulator adjusting knob completelycounterclockwise.

10. Modify your circuit as shown in Figure 4-5.


8

DCV 1

DCV 2

Figure 4-5. Schematic Diagram of a Circuit Using a Long Tubing Line.

11. Open the shutoff valves and set the pressure regulator at 100 kPa (or15 psi) on the regulated Pressure Gauge.

12. Push down the button on the directional control valve while observing thetime taken by the rod to extend fully. Estimate the time taken by the rod toextend fully and record your result in Table 4-1. Repeat your observationthree times, then calculate the mean value.

13. Close the shutoff valves and turn the regulator adjusting knob completelycounterclockwise.

14. Modify your circuit as shown in Figure 4-6.

Figure 4-6. Schematic Diagram of an Indirect Control Circuit.


9

15. Open the main shutoff valve and set the pressure regulator at 100 kPa (or15 psi) on the regulated Pressure Gauge.

16. Push down the button on the directional control valve DCV2 to set thespool in the pilot-operated valve as illustrated in Figure 4-6. The piston rodshould be retracted.

17. Push down the button on the directional control valve DCV1 whileobserving the time taken by the rod to extend fully. Estimate the time takenby the rod to extend fully and record your result in Table 4-1. Repeat yourobservation three times, then calculate the mean value.

READINGEXTENSION TIME OF THE PISTON ROD

DIRECT CONTROL INDIRECT CONTROL

First Reading

Second Reading

Third Reading

Mean Value

Table 4-1. Extension Time of The Piston Rod

18. Compare the results shown in Table 4-1. Does the rod extend faster whenthe valve is controlled indirectly?

Yes No

19. Since the same Long Line was first used to power the cylinder and then topilot the directional control valve, what can you conclude?

20. Does the piston rod retract when the button on the directional control valveDCV1 is released? Explain why.


10

21. On the Conditioning Unit, close the shutoff valves and turn the regulatoradjusting knob completely counterclockwise. You should read 0 kPa (or0 psi) on the regulated Pressure Gauge.

22. Disconnect and store all tubing and components.

CONCLUSION

In this exercise, you verified the operation of a 4-way, 5-port, 2-positionpilot-operated directional control valve. You have seen that the piloted valvesupplied with your trainer can be used to remotely control linear and rotaryactuators in two directions.

You have seen that when a pilot port is maintained pressurized, priority ismaintained on that position although the other port becomes pressurized.

You have seen that pilots require minimum pressures and volumes to shift againstthe working pressures, minimizing delays caused by compressibility of air andfriction in long tubing lines.

REVIEW QUESTIONS

1. What is the greatest advantage of pilot-operated valves over manuallyoperated valves?

a. They require a high pressure to operate.b. They permit the remote actuation of large valves.c. They can be made smaller than other valves.d. They can be made larger than other valves.

2. What is the purpose of the 4-way piloted-operated valves?

a. To control linear actuators in one direction remotely.b. To control rotary actuators in one direction remotely.c. To control linear and rotary actuators in one direction remotely.d. To control linear and rotary actuators in two directions remotely.

3. What is the main difference between pilot-operated and direct-operated controlvalves?

a. Pilot-operated control valves are smaller.b. The way their spools are shifted.c. Pilot-operated control valves can work in both directions.d. Pilot-operated control valves cannot be spring return.


11

4. What is the purpose of the manual override on a pilot-operated valve?

a. To bleed off excess compressed air.b. To reverse the direction of the valve.c. To reverse pilot operation.d. To manually duplicate the operation of the valve.

5. Give the reason why double pilot-operated valves can memorize a position?

a. They need a pilot signal to shift the spool.b. They do not need a pilot signal to shift the spool.c. It is a characteristic of pilot-operated valve.d. Because they are remotely-controlled.

Sample Exercise

from

Electrical Control of

Pneumatic Systems

15

DV3

SOL-B

DV1

SOL-A

DV2

CYLINDER

CHECKVALVE

P1

P2

Exercise 3-1

Basic Memory and Priority Electropneumatic Circuits

EXERCISE OBJECTIVE

• To show how a directional valve can memorize a signal and maintain a position;• To demonstrate how to lock and unlock electropneumatic circuits;• To describe the function and operation of limit switches.

DISCUSSION

As seen in Exercise 4-1 in the Pneumatics Fundamentals manual, the lack of returnsprings, in double-air-piloted directional valves, allows to maintain or memorize thespool position without maintaining pilot pressure. However, when a pilot port ismaintained pressurized, priority is maintained on that position although the otherport becomes pressurized. The circuit must exhaust one pilot before the other pilotcan shift the spool.

In the circuit shown in Figure 3-1, directional valve DV2 acts as a memory controlvalve. Before operating the circuit, the spool position of DV2 is unknown, and itcorresponds to the last position the valve was used. To maintain the cylinderretracted when starting the circuit, the spool must be positioned by a manualoverride or by a pilot signal.

Figure 3-1. Memory and Priority Basic Circuit.

When solenoid SOL-A of directional valve DV1 is energized, the spool of the valveshifts and compressed air flows to pilot P1 of DV2. This causes the spool of DV2to shift and the cylinder to extend. The lack of return springs causes DV2 to


16

memorize and maintain its position so the cylinder will continue to extend and willremain extended although SOL-A is de-energized.

When solenoid DV1-SOL-B is energized to retract the cylinder, compressed airflows to pilot P2 of DV2 but the spool of DV2 does not shift. It is blocked bycompressed air which is trapped by the check valve in the pilot line of P1.Directional valve DV3 is then used to bleed the compressed air.

Since the retraction of the cylinder must be confirmed by a second command(DV3), this type of circuit is often used to prevent unwanted operations. Theoperation of this air-locked circuit is similar to the interlock circuit shown inFigure 2-19 in Exercise 2-3 where it is necessary to depress the STOP pushbutton,as a second command, before energizing the opposite solenoid.

Limit Switches

A command can also be confirmed using the electrical signal provided by sensingdevices which detect the position of the cylinder rod. As an example, the retractioncommand of the cylinder rod shown in Figure 3-2, could not be executed by PB2if the rod is not fully extended and its position confirmed by the limit switch LS2.When LS2 is mechanically activated by the presence of the rod, its NO contactgoes closed, and it is therefore possible to energize CR2 using PB2. Energizingrelay coil CR2 causes NC contact CR2-A to go open and relay coil CR1 to de-energize.

Limit switches are used extensively in industrial pneumatic equipment. They arereliable, small in size, simple to use, and generally cheaper than the other types ofswitches. A limit switch consists of an actuator and one or more sets of NO andNC contacts. It is activated when a moving part, such as a cylinder rod or machinemember, strikes the actuating mechanism, shifting the contacts to their activatedstate.

Figure 3-3 shows the limit switch assembly supplied with your trainer. Each switchhas a roller-type actuator and a set of SPDT contacts. When the cylinder tip travelsacross one of the switches, it pushes against the roller, depressing the lever arm.The lever arm acts on an internal plunger, causing the SPDT contacts to activate.The NO contact goes closed while the NC contact goes open. When the cylindertip moves away from the roller actuator, a spring returns the lever arm and thecontacts to their normal condition.


17

LADDER DIAGRAM

(+)

PB1

(−)

L1

1 CR1

CR1-A DV1-SOL-A

PNEUMATIC DIAGRAM

CR2-A

CR2-B

PB2

CR1-B DV1-SOL-B

L2

CR2

DV1

SOL-B

SOL-A

LS1

LS22

LS1 LS2

Figure 3-2. Electropneumatic Circuit Using Limit Switches.


18

LADDER DIAGRAM SYMBOL

COMMON TERMINAL

PNEUMATIC DIAGRAM SYMBOL

ROLLER ACTUATOR

LEVER ARMSPRING

NO TERMINAL

NC TERMINAL

Figure 3-3. Limit Switch with Roller Arm Actuator.

Limit switches are often available as a multiple-switch assembly with two or morelimit switches mounted on the same supporting frame. The two limit switchessupplied with your trainer are mounted on the same supporting frame. This designis ideal for situations which require two switches mounted side by side.

REFERENCE MATERIAL

Procedure Summary

In the first part of the exercise, you will test the operation of a basic memory andpriority circuit using a double-air-piloted directional valve.

In the second part, you will test a locking circuit, using compressed air trapped ina pilot line by a check valve, to maintain the pilot port pressurized and to maintainthe position of the valve.

In the third part, you will learn how to mount the limit switches supplied with yourtrainer.

In the last part of the exercise, you will test a priority locking circuit using limitswitches to confirm the position of the cylinder rod.

EQUIPMENT REQUIRED



19

PNEUMATIC DIAGRAM

FCV 2

FCV 1

DV1

DV2

DV3

PROCEDURE

Basic Memory and Priority Circuit

1. Connect the circuit shown in Figure 3-4. Screw a tip to the cylinder rod.

Figure 3-4. Schematic Diagram of a Memory Circuit.

2. Verify the status of the trainer according to the procedure given inAppendix F.

3. Close the flow control valves by turning the control knobs fully clockwise.Then open each valve by turning the knobs two turns counterclockwise.Refer to the mark on the knobs to help you set the correct position.

Note: The flow control valves are used to control the extensionand retraction speeds of the cylinder.

4. Will the cylinder rod extend, or retract, when compressed air will be appliedto the circuit? Explain.


20

5. On the conditioning unit, open the main shutoff valve and the branchshutoff valves at the manifold. Set the pressure regulator at 400 kPa(or 60 psi) on the regulated pressure gauge.

6. If necessary, depress the control button of directional valve DV2 to retractthe cylinder rod.

7. Depress the control button of directional valve DV1. Does the cylinder rodextend?

8. Depress the control button of directional valve DV1 and hold this position.With your other hand, depress the control button of DV2. Does the cylinderrod retract? Explain why.

9. Does the operation of the circuit confirm that priority can be maintained ona position when the pilot port remains pressurized?

Yes No

10. On the conditioning unit, close the shutoff valves, and turn the regulatoradjusting knob completely counterclockwise.

Priority Locking Circuit

11. Connect the priority locking circuit shown in Figure 3-5. Use a closed flowcontrol valve (turn the control knob fully clockwise) as check valve in thepilot line of P1. Ensure that the cylinder rod is retracted. Screw a tip to thecylinder rod.


21

LADDER DIAGRAM

(+)

PB1

(−)

L1

1 CR1

DV1-SOL-A

PNEUMATIC DIAGRAM

CR2-A

PB2CR1-A

DV1-SOL-B

L2

CR22

DV3

SOL-B

DV1

SOL-A

DV2

CYLINDER

CV1

P1

P2FCV1

Figure 3-5. Schematic Diagram of a Priority Locking Circuit.

12. Close flow control valve FCV1 by turning the control knob fully clockwise.Then open the valve two turns counterclockwise.

13. On the conditioning unit, open the shutoff valves, and set the pressure at400 kPa (or 60 psi).


22

14. Turn on the DC power supply.

15. If the cylinder rod extends when applying pressure in the circuit, depresssimultaneously PB2 and the control button of directional valve DV3 toretract the cylinder rod.

16. Depress pushbutton PB1. Does the cylinder rod extend?

Yes No

17. Depress pushbutton PB2. Does the cylinder rod retract? If not, explain.

18. Depress simultaneously pushbutton PB2 and the control button ofdirectional valve DV3. Does the cylinder rod retract?

Yes No

19. Does the operation of the circuit confirm that priority is maintained on aposition when the pilot port remains pressurized?

Yes No

20. Explain how the circuit will operate if the check valve is removed from thecircuit.


22. Turn off the DC power supply.


23

Mounting of the Limit Switch Assembly

23. Remove all components except the double-acting cylinder from your worksurface.

24. Mount the limit switch assembly as indicated in the following steps:

• Screw the cylinder tip onto the rod end of the cylinder.

• Manually extend the cylinder rod completely.

• Clamp the limit switch assembly along the cylinder rod as shown inFigure 3-6.

• Loosen the limit switch positioning screws. Position the switches sideby side at the center of the support bracket, as shown in Figure 3-6 (a).Tighten the limit-switch positioning screws.

• Loosen the support-bracket positioning screws until you are able toslide the bracket over the mounting base as shown in Figure 3-6 (b).Adjust the position of the bracket so that the switches are activatedwhen the cylinder tip pushes against the switch arm and deactivatedwhen the cylinder tip releases the switch arm. To test this out,manually extend and retract the cylinder rod, and listen for the "click".Then, tighten the support-bracket positioning screws on the mountingbase.

• Loosen the positioning screw on each limit switch. Adjust thepositioning of the switches so that they are activated when the cylinderrod is fully extended and fully retracted, as shown in Figure 3-6 (c). Totest this out, manually extend and retract the cylinder rod, and listen forthe click. Then, tighten the limit-switch positioning screws.

• Retract the cylinder rod completely, as shown in Figure 3-6 (d).

25. Your limit switch assembly is now set to detect the fully-extended and fully-retracted positions of the cylinder rod.


24

TIP

POSITIONING SCREWSLIMIT-SWITCH

BRACKETSUPPORT

SUPPORT BRACKETPOSITIONING SCREWS

BASEMOUNTING

DEACTIVATED ACTIVATED

(a) (b)

(c) (d)

FULLYRETRACTED EXTENDED

FULLY ACTIVATED

Figure 3-6. Mounting of the Limit Switch Assembly.

Priority Locking Circuit Using Limit Switches

26. Connect the circuit shown in Figure 3-7. As you do this, be careful not tomodify the mounting of the limit switches LS1 and LS2.


25

LADDER DIAGRAM

(+)

PB1

(−)

L1

1 CR1

CR1-A DV1-SOL-A

PNEUMATIC DIAGRAM

CR2-A

CR2-B

PB2

CR1-B DV1-SOL-B

L2

CR2

DV1

SOL-B

SOL-A

LS1

LS22

LS1 LS2

FCV1

FCV2

Figure 3-7. Schematic Diagram of a Priority Locking Circuit Using Limit Switches.

27. Close the flow control valves by turning the control knobs fully clockwise.Then open each valve by turning the knobs two turns counterclockwise.Refer to the mark on the knobs to help you set the correct position.




26

30. Depress pushbutton PB1. Does the cylinder rod extend?

Yes No

31. Depress pushbutton PB2. Does the cylinder rod retract?

Yes No

32. Loosen the positioning screw of the limit switch which detects the positionfully extended and pull up the limit switch.

33. Depress pushbutton PB1 to extend the cylinder rod, then depresspushbutton PB2. Does the cylinder rod retract? If not, explain by referringto the ladder diagram.

34. Does the operation of the circuit confirm that the limit switch must confirmthe position of the cylinder rod to allow the spool of directional valve DV1to be shifted?

Yes No


36. Turn off the DC power supply.

37. Disconnect and store all leads and components.

CONCLUSION

In this exercise, you learned how priority can be determined and maintained in apneumatic circuit, using a double-air-piloted directional valve.

You tested a locking circuit using compressed air trapped in a pilot line. You sawthat the circuit must exhaust one pilot before the other pilot can shift the spool. Yousaw that the circuit remains locked even though compressed air supply falls.


27

You learned how to mount the limit switch assembly supplied with your trainer.

You tested an interlock circuit which needs a confirmation command, supplied bya limit switch, to allow a new sequence to be started.

REVIEW QUESTIONS

1. What is the purpose of directional valve DV3 in Figure 3-5?

2. What care should be take before shifting the spool of a double-air-piloteddirectional valve?

3. What is the purpose of a limit switch in an electropneumatic circuit?

4. What characteristic of a double-air-piloted directional valve allows to maintainor memorize the spool position without maintaining pilot pressure?

5. What are the purposes of contacts CR1-A and CR2-A in the ladder diagram ofFigure 3-7?

Sample Exercise

from

Pneumatics Applications – PLC

31

Exercise 6

Counting of Pneumatic Actuator Cycles

EXERCISE OBJECTIVE

• To connect and test a PLC-controlled pneumatic system that makes a motorrotate 200 turns and then reciprocates a cylinder 5 times.

DISCUSSION

Counting of pneumatic actuator cycles is required when a portion of a system mustbe activated or deactivated after an actuator has completed a definite number ofcycles. A typical application is an automated packing machine that stacks andcounts production items into groups. The usual method is for a cylinder tocontinuously extend and retract, picking and stacking one item on each cycle, andfor a counter to count the number of cycles which have been completed by thecylinder. When the required count is reached, a switching signal causes anothercylinder to push the stack away.

Counting of pneumatic actuator cycles is also required for machine maintenancescheduling. The PLC keeps track of the number of items the machine manufacturesto determine when a part should be replaced.

The PLC counter instructions are ideally suited for counting the number of cyclescompleted by an actuator. They allow monitoring of automatic production machinesat high efficiency rates.

Procedure Summary

In this exercise, you will connect a PLC-controlled pneumatic system that makesa motor rotate 200 turns and then reciprocates a cylinder 5 times.

EQUIPMENT REQUIRED



32

PROCEDURE

1. Connect the PLC-controlled pneumatic system shown in Figure 6-1. Notethat the long line device is used to decrease the retraction speed of thecylinder rod.


33

SUPPLYPOWER

DC

PLC

INPUTCOMMON

+

PE1, NO

PB2, NC

PB1, NO TERMINALSINPUTPLC

INPUTPOWERRELAY

_

+

OUTPUTTERMINALS

PLC

PLC CONTROL SECTION

DV1-SOL-A

DV2-SOL-A

0 V (COM)

6

11

10

8

9

7

5

4

3

2

1

0

6

7

5

4

3

2

1

0

24 V dc

DV2-SOL-B

PX1, NO

PX2, NO

PNEUMATIC DIAGRAM

PX1

FCV2

PE1

DV2

SOL-A

SOL-A

FCV1

DV1

SOL-B

MOTORBI-DIRECTIONAL

PX2

Figure 6-1. PLC-controlled pneumatic system.


34

2. Position the pneumatic motor so that it is perpendicular to the photoelectricswitch at a distance of 10 cm (4 in) between the switch and the motor axis(2 rows of perforations). The photoelectric switch beam must be pointingin the direction of the white sticker on the motor shaft.

3. Mount magnetic proximity switches PX1 and PX2 so that they are activatedwhen the cylinder rod is fully extended and fully retracted, as shown inFigure 6-1.

4. Enter the PLC ladder program shown in Figure 6-2, using the instructionaddress format appropriate to your model of PLC.

5. Download your program to the PLC. Place the PLC in Run mode.

6. Verify the status of the trainer according to the procedure given inAppendix F.

7. Open flow control valve FCV1 by turning the control knob fullycounterclockwise. Close flow control valve FCV2 by turning the controlknob fully clockwise. Then open the valve by turning the knob two turnscounterclockwise.




35

RUNG 1

RUNG 0

(+) (-)

PB1I:00

PB2I:01

RUNG 11 END

WORKBIT 0B:0

RUNG 2

DV1-SOL-AO:00OUT

BITC:1

C:1 COMPLETION

RUNG 4

RUNG 3

B:0BIT 0

UP COUNTER

C:0

RESET

COUNT

n = 200

B:0BIT 0

WORK

LB:1

LATCH B:1I:04PX2

RUNG 5

START

LB:0

LATCH B:0

B:0UNLATCH B:0

U

STOP

I:04PX2

C:0 COMPLETION

C:0BIT

PE1I:02

C:0 COMPLETION

C:0BIT

WORK

RUNG 6I:03PX1

B:1UNLATCH B:1

U

RUNG 7

WORKBITC:0 B:1

BIT 1O:01

DV2-SOL-A

OUT

C:0 COMPLETION

RESET

UP COUNTER

COUNT

C:1

n = 5

RUNG 9

RUNG 8

PX1

WORKBIT 0B:0

I:03

RUNG 10

WORKBIT 0B:0 I:04

PX2

BITC:0 COMPLETION

C:0OUTO:02

DV2-SOL-BB:1

BIT 1WORK

Figure 6-2. PLC ladder program.


36

10. Test the operation of the system, using the following verification steps:

• Momentarily depress the START pushbutton PB1. The motor shouldstart rotating, while the cylinder rod should remain immobile;

Note: If the LED on the photoelectric switch seems to skip,reduce the rotation speed of the motor by decreasing the air flowwith FCV1.

• Monitor PLC counter instruction C:0 in ladder rung 3. When thecounter accumulated value reaches 200, the motor should stop, whilethe cylinder rod should start to reciprocate (extend and retract);

• When the cylinder rod has reciprocated 5 times, it should stop in thefully retracted position, allowing a new cycle to be initiated.

If the operation of the system is not as indicated, check the circuitconnections and the PLC program. Perform the required modifications,then verify that the system operates correctly.

11. Cycle the system a few times and monitor the PLC program as it isexecuted. What causes the motor to start rotating when PB1 is depressed?Explain by referring to the PLC ladder program shown in Figure 6-2.

12. What causes counter instruction C:0 in rung 3 to increment its accumulatedvalue when the motor rotates? Explain.


37

13. What causes the motor to stop and the cylinder rod to start reciprocatingwhen the accumulated value of counter instruction C:0 reaches 200?Explain.

14. What causes counter instruction C:1 in rung 8 to increment its accumulatedvalue every time the cylinder rod becomes fully extended?

15. What causes the cylinder rod to stop in the fully retracted position after ithas reciprocated 5 times? Explain.

16. Start the system by depressing PB1, then depress PB2 while the motor isrotating. Does the motor stop immediately when PB2 is depressed? Why?


38

PB1 (START)

DV1-SOL-A

PE1

DV2-SOL-A

PX1

DV2-SOL-B

ROD MOTION

t

t

t

t

t

t

tPX2

MOTOR

17. Start the system by depressing PB1, then depress PB2 while the cylinderrod is extending and in mid-stroke. Does the cylinder rod return to the fullyretracted position before it stops? Why?

18. Use Figure 6-3 to draw the timing diagram of the system.

Figure 6-3. Timing diagram.

19. Modify your PLC program so that the system will operate as follows:

Depressing the START pushbutton causes the cylinder rod to startreciprocating, while the motor remains stopped.When the rod has reciprocated 5 times, it stops, while the motor startsto rotate.When the motor has rotated 200 turns, it stops and the systembecomes ready for a new cycle.


39

(+) (-)

Use Figure 6-4 to draw the modified PLC program. Enter your program andtest system operation.

Figure 6-4. New modified program.

20. Turn off the PLC, the DC power supply and the host computer, if any.



40


CONCLUSION

In this exercise, you connected a PLC-controlled pneumatic system that makes amotor rotate 200 turns and then reciprocates a cylinder 5 times.

You saw that the PLC counter instruction is incremented by false-to-true rungtransitions of the counter rung. These rung transitions are caused by eventsoccurring in the system, such as a cylinder piston traveling past a magneticproximity switch or a motor flywheel activating a photoelectric switch. After adefinite number of events have occurred, the counter completion bit turns on, whichactivates or deactivates a directional valve solenoid.

REVIEW QUESTIONS

1. When is counting of pneumatic actuator cycles required?

2. Describe a typical pneumatic application where counting of actuator cycles isrequired.

3. How can the PLC counter instruction be used to activate a directional valvesolenoid after a cylinder has reciprocated a definite number of times?


41

4. In the PLC ladder program of Figure 6-2, what purpose is served by NC contactinstruction C:0 in rung 3?

5. In the PLC ladder program of Figure 6-2, what purpose is served by NC contactinstruction B:0 in rung 4?

Sample Exercise

from

Servo/Proportional Control

of Pneumatic Systems

45

Exercise 5

Closed-Loop Position Control,Proportional-Plus-Integral Mode

EXERCISE OBJECTIVE

To describe the integral control mode;To define the terms integral gain and overshoot;To describe the advantages and disadvantages of integral control;To describe the proportional-plus-integral control mode.

DISCUSSION

The Integral (I) Control mode

While the proportional control mode looks at the "present" value of the processerror, the integral (I) control mode regards the "past history" of the error bycontinuously integrating it until it is eliminated. Thus, the integral control modeautomatically reduces the residual error to zero for any load change within thelimitations of the system design, thereby eliminating the need for manual resetassociated with the proportional control mode.

Figure 5-1 shows the simplified diagram of a controller operating in the integralmode:

The "error detector" compares the measured position to the setpoint andproduces an error signal equal to the difference between the two;

The "integral amplifier", integrates the error to produce the controller outputsignal.


46

FEEDBACKSIGNAL

SETPOINTCONTROLLEROUTPUT (C )

AMPLIFIER

+

−

ERROR (E )

ERRORDETECTOR

P

INTEGRAL CONTROLLER

INTEGRAL

OOEKtI 0

tCdt +P (t )0

K0I t

t(DESIREDPOSITION)

(MEASUREDPOSITION)

Figure 5-1. Simplified diagram of a controller operating in the integral (I) mode.

The output of the controller at any specified point in time, is given by:

where CO (t) is the controller output at a specified time;KI is the integral gain;EP is the error at a specified time;

CO (t0) is the controller output at the time the observation starts (t = 0).

The controller output signal is at all times proportional to the time integral of theerror, hence the name integral mode.

Figure 5-2 shows an example of what happens to the signal at the output of anintegral controller when the error is positive, negative, and null. The controller is inthe open-loop mode. As you can see, when the error is positive, the controlleroutput increases. When the error is negative, the controller output decreases.When the error is null, the controller output remains at the same level.


47

ERROR

OUTPUT

CONTROLLER

0

(+)

(−)

TIME

(−)

0

TIME

(+)

NULL

NEGATIVE

POSITIVEVOLTAGE

VOLTAGE

Figure 5-2. Controller output signals with positive, negative and null errors.

Integral gain

Figure 5-3 shows the output signal of an integral controller for different integral gainsettings when the error changes suddenly. The controller is in the open-loop mode.The integral action causes the error signal to be transformed into a graduallychanging signal at the controller output. The greater the integral gain KI, the greaterthe rate of change of the controller output.

The integral gain is expressed in number of repeats per minute (rpt/min). Itcorresponds to the number of times the magnitude of the error is duplicated at thecontroller output in a period of 1 minute:

With a gain of 5 rpt/min, the 1.0-V error magnitude is duplicated5 times in 1 min. This means the controller output will be 5.0 V after 1 min (seeFigure 5-3);

With a gain of 10 rpt/min, the 1.0-V error magnitude is duplicated 10 timesin 1 min. This means the controller output will be 10.0 V after 1 min.

With a gain of 50 rpt/min, the 1.0-V error magnitude is duplicated 50 timesin 1 min. This means the controller output will be 10.0 V after 0.2 min.


48

ERROR

TIME (min)1 2t 0

1

0

2

3

4

5

6

7

8

9

10

VOLTAGE (V) CONTROLLER OUTPUT K = 50 rpt/minI

CONTROLLER OUTPUTK = 10 rpt/minI

CONTROLLER OUTPUTK = 5 rpt/minI

0.2

Figure 5-3. Output signal of an integral controller in the open-loop mode when the error changessuddenly.

Now suppose the integral controller is placed in the "closed-loop mode" in order tocontrol a process. Figure 5-4 shows the effect of increasing the integral gain on theresponse of the controlled variable to a sudden change in setpoint. The higher theintegral gain is, the shorter the time required to eliminate the error will be.

With a low integral gain as in waveform (a), the controlled variable is broughtback to the new setpoint relatively slowly but with no overshoots. The systemresponse is said to be “damped”. A damped response may be desirable insome applications.

Increasing the integral gain as in waveforms (b) reduces the time required toeliminate the error, but may also cause the controlled variable to overshoot thenew setpoint before it stabilizes. However, small overshoots are usuallytolerated in most applications.

Further increasing the integral gain results in higher overshoots, as inwaveform (c). Worse still, the time required to eliminate the error, instead ofdecreasing, actually gets longer because the controlled variable performs anumber of oscillations about the setpoint before it stabilizes.

At very high integral gains, the system may even start to oscillate without beingable to return to equilibrium, as in waveform (d).


49

TIME

(a) DAMPED (b) CORRECT (c) UNDERDAMPED (d) OSCILLATORY

OVERSHOOT

K = 30 rpt/minI IK = 45 rpt/min K = 60 rpt/minI IK = 75 rpt/min K = 150 rpt/minI IK = 300 rpt/min

MEASURED

SETPOINT

VOLTAGE

VARIABLE

Figure 5-4. Effect of increasing the integral gain on the step response of a controlled variable.

The ability to reestablish a zero error is a unique characteristic of the integralcontrol mode and is found in no other mode. However, the integral mode isnormally not used alone, because of its relatively slow response at low integralgains, and because of the overshoots and increased risks of oscillation at higherintegral gains. Instead, the integral mode is combined with the proportional modeto form the “proportional-plus-integral” control mode.

Proportional-Plus-Integral (P.I.) Control mode

The proportional-plus-integral control mode combines the fast transient responseof proportional control with the zero residual error characteristic of integral control.It looks at the current value of the error and the integral of the error over a recenttime interval to determine how much of a correction to apply, and also for how long.

Figure 5-5 shows the simplified diagram of a controller operating in theproportional-plus-integral mode. The error produced by the error detector is firstamplified by a factor KP by the proportional amplifier. The proportionally amplifiederror (EP KP) is then time integrated by the integral amplifier. Finally, the outputsignals of the proportional and integral amplifiers are added at a summing point toproduce the controller output signal.


50

FEEDBACKSIGNAL

SETPOINT

K

CONTROLLEROUTPUT (C )

AMPLIFIER

+

−

ERROR (E )

ERRORDETECTOR

P

PROPORTIONAL-PLUS-INTEGRAL CONTROLLER

PROPORTIONAL

P

EP dt + OKI 0t

tCO (t )0

0(t )OCt

t 0IK

0t

t dt +PEI

INTEGRALAMPLIFIER

K

PKE P.

KP.

POINTSUMMING

+

+ .PE K P P+ K .(DESIRED

POSITION)

(MEASUREDPOSITION)

Figure 5-5. Simplified diagram of a controller operating in the proportional-plus-integral (P.I.)mode.

The controller output at any specified time, t, is given by:

where CO (t) is the controller output at a specified time;EP is the error at a specified time;KP is the proportional gain;KI is the integral gain;

CO (t0) is the controller output at the time the observation starts (t = 0).

In this equation, the term "EP × KP" describes the proportional action of thecontroller, while the rest of the equation describes the integral action of thecontroller, starting at time t = 0. Thus, the controller output is not only proportionalto the error, but also to the time integral of the error.

Figure 5-6 shows what happens in a proportional-plus-integral control system whenthe error changes suddenly, due to a sudden increase in setpoint or actuator load.The proportional action quickly responds to the sudden change in error, while theintegral action integrates the error until it becomes null. The higher the integral andproportional gains are, the faster the error will be eliminated. However, these gainsmust be kept low enough to prevent excessive overshoots, and to ensure systemstability.


51

ERROR

0

(+)

(−)

TIME

(−)

0TIME

(+)

C (t )

FINAL VALUE OF C O

0O

CONTROLLER

OUTPUT (C )O

INTEGRALACTION

PROPORTIONALACTION

VOLTAGE

VOLTAGE

Figure 5-6. Example of what happens in a proportional-plus-integral control system when theerror changes suddenly.

A problem related to the use of integral control is the "reset windup". Reset windupoccurs when the error remains large during a long interval of time, causing thecontroller output to increase to the upper saturation limit (positive error) or decreaseto the lower saturation limit (negative error). Afterwards, the controller cannot returnto normal operation until the error reverses polarity, resulting in high overshoots ofthe controlled variable. For that reason, controllers usually have an "anti-reset"function that turns off the integral action as soon as the controller output reachesthe upper or lower saturation limit, thus minimizing overshoots of the actuatorposition.

Comparison of the Proportional, Integral, and Proportional-Plus-IntegralControl Modes

Figure 5-7 shows an example of the response of a typical process to a step changein setpoint with proportional, integral, and proportional-plus-integral control modes :

The proportional mode reacts much faster than the integral mode. However, theproportional mode results in a residual error between the actual rod positionand the new setpoint;


52

SETPOINTAND

MEASUREDPOSITION

TIME

SETPOINT

PROPORTIONAL-PLUS-INTEGRAL

PROPORTIONAL

VOLTAGE

INTEGRAL

The integral mode eliminates the residual error, but it reacts slower than theproportional mode, and it requires a longer time to reach the final value;

The proportional-plus-integral mode reacts much faster than the integral mode,and it eliminates the residual error of the proportional mode. However, theaddition of integral action to proportional action increases the overshoot andstabilization time.

Figure 5-7. Response to a step change in setpoint with proportional, integral, and proportional-plus-integral control modes.

Procedure summary

In the first part of the exercise, Familiarization with the Proportional-Plus-Integral(P.I.) Controller, you will familiarize yourself with the components and operation ofa proportional-plus-integral controller.

In the second part of the exercise, Proportional-Plus-Integral (P.I.) Control ofCylinder Rod Position, you will study proportional-plus-integral control of cylinderrod position. You will observe the effect of increasing the integral and proportionalgains on the response of the rod position (feedback voltage) when the setpointchanges suddenly.

EQUIPMENT REQUIRED

Refer to the Equipment Utilization Chart, in Appendix A of the manual, to obtain thelist of equipment required to perform this exercise.

PROCEDURE

Familiarization with the Proportional-Plus-Integral (P.I.) Controller

1. Connect the circuit shown in Figure 5-8. In this system, a P.I. controller isbuilt with the ERROR DETECTOR, the PROPORTIONAL and INTEGRALAMPLIFIERS, the SUMMING POINT, and the LIMITER of the PIDController.


53

222

PROPORTIONAL

P.I.D. AMPLIFIERS

P[ ]

LOW

MIN.

HIGH

POWER

LOWPASS FILTER

0.02 Hz 2 Hz

fc

−10 V - +10 V

DERIVATIVEINTEGRAL[ ]I

MIN.MAX.

[ ]D

MIN.MAX.

ANTI-RESET

MAX.

SUMMING POINT

I

D

GAINS

P

MIN.

INTEGRATOR

++

+

MAX.

MIN.

LOWER

MAX.MIN.

UPPER

MAX.

LIMITS

LIMITER

24 V

1

CR

ACR

BCR

B

−10 V - +10 V

1

SETPOINTS

S1A

RAMP GENERATOR

P.I.D. CONTROLLER

1

MIN. MAX.

++

−

ERROR DETECTOR

24-V DC POWER SUPPLY

1

+ −

24 V − 2.4 A 2

POSITION TRANSDUCER

2

24 V

3

0-5 V

0-10 V

0-5 V

0-10 V

0-5 V

0-10 V

2-33 Hz

24 V

+

24 V

E

−

N

1 V / 1000 r/min

f

S

0-24 V

0-5 V

E

E

E

E

SZ

SIGNAL CONDITIONERS

−+24 V

0-10 V

0-5 V

0-10 V

SERVO CONTROL VALVE

1

POSITIONTRANSDUCER

CONNECTION DIAGRAM

PNEUMATIC DIAGRAM

PG2REGULATORPR2

PRESSURE

SERVO CONTROLVALVE

Note: Do not connect the 0-10 V output of the [0-24 V] E/Econverter to the negative input of the ERROR DETECTOR at thistime. This will be done later in the exercise.

Figure 5-8. Proportional-plus-integral (P.I.) control of cylinder rod position.


54

2. Make the following settings on the PID Controller:

PROPORTIONAL [P] GAIN range . . . . . . . . . . . . . . . . . . . . . . . LOWPROPORTIONAL [P] GAIN . . . . . . . . . . . . . . . . . . . . . . . . . . . . MIN.INTEGRAL [I] GAIN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . MIN.INTEGRATOR ANTI-RESET . . . . . . . . . . . . . . . . . . . . . . . . . . . . . OLOWER LIMIT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ½ of MAX.UPPER LIMIT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ½ of MAX.

3. Turn on the DC Power Supply and PID Controller.

4. On the PID Controller, set the SETPOINT potentiometers 1 and 2 toobtain 0.25 V and 0.25 V, respectively, at the SETPOINT output 1. Thenselect the SETPOINT potentiometer 1.

5. On the PID Controller, set the PROPORTIONAL GAIN to obtain 0.25 V atthe proportional amplifier output. This will set the gain of the proportionalamplifier at about 1 and will apply a positive voltage of 0.25 V at theintegral amplifier input.

6. Monitor the DC voltage at the integral amplifier output. You should observethat the voltage slowly increases because the voltage, at the integralamplifier input, is positive.

You should also observe that the voltage stops increasing when it reachesthe positive saturation level of the integral amplifier (about 14 V). Is thisyour observation?

Yes No

7. Set the INTEGRATOR ANTI-RESET switch at I. Is the integral amplifieroutput voltage still 14.0 V or is it now limited to the UPPER LIMITpotentiometer setting of approximately 7 V?

8. Adjust the UPPER LIMIT potentiometer in order to obtain a stable voltageof 10.0 V at the integral amplifier output. This sets the upper limit at 10.0 V.

9. Set the INTEGRATOR ANTI-RESET switch at O.


55

10. Select SETPOINT potentiometer 2. This will apply a negative voltage ofabout 0.25 V at the integral amplifier input.

11. Monitor the DC voltage at the integral amplifier output. You should observethat the voltage slowly decreases because the voltage, at the integralamplifier input, is negative.

You should also observe that the voltage stops decreasing when it reachesthe negative saturation level of the integral amplifier (about 14 V). Is thisyour observation?

Yes No

12. Set the INTEGRATOR ANTI-RESET switch at I. Is the integral amplifieroutput voltage still 14.0 V or is it now limited to the LOWER LIMITpotentiometer setting of approximately 7 V?

13. Set the LOWER LIMIT potentiometer to MIN. This sets the lower limitat 0.0 V.

14. Select SETPOINT potentiometer 1 and set the potentiometer to obtain anull voltage (0.0 V) at the integral amplifier input.

Now monitor the voltage at the output of the integral amplifier. You shouldobserve that the voltage is constant (or increasing very slowly) at someintermediate value. This shows that the output voltage of an integralamplifier is constant when the input voltage is null. Is this yourobservation?

Yes No

15. Set the SETPOINT potentiometer 1 to obtain 1.0 V at the integral amplifierinput.

16. Now observe the effect of increasing the INTEGRAL GAIN on the rate ofchange of the integral amplifier output voltage.

With the DC voltmeter connected to the integral amplifier output, set theINTEGRAL GAIN potentiometer at ½ of MAX., then at ¾ of MAX., and thento MAX. For each setting, alternately select SETPOINT potentiometers 1and 2.


56

You should observe that the rate of change of the integral amplifier outputvoltage increases as the INTEGRAL GAIN is increased. Is this yourobservation?

Yes No

17. Select SETPOINT potentiometer 1 and set the INTEGRAL GAIN to MAX.

18. Measure the output voltages of the proportional and integral amplifiers.These voltages should be about 1 V, and 10 V, respectively.

Now measure the voltage at the SUMMING POINT output. Is this voltageabout 11 V? Explain.

19. Measure the voltage at the LIMITER output. This voltage should be limitedto the UPPER LIMIT potentiometer current setting of 10.0 V. Is this yourobservation?

Yes No

Proportional-Plus-Integral (P.I.) Control of Cylinder Rod Position

20. On the PID Controller, set the SETPOINT potentiometers 1 and 2, thePROPORTIONAL GAIN and the INTEGRAL GAIN as follow:

Set the SETPOINT potentiometers 1 and 2 to obtain 0.0 V and 5.0 Vrespectively, at the SETPOINT output 1;

Select the SETPOINT potentiometer 2, and set the PROPORTIONALGAIN to obtain 5.0 V at the proportional amplifier output.

Set the INTEGRAL GAIN at ½ of MAX.

21. Verify the status of the trainer according to the procedure given inAppendix B.

22. On the Conditioning Unit, open the main shutoff valve and the requiredbranch shutoff valves at the manifold.

Set the main pressure regulator to obtain 630 kPa (90 psi) on the regulatedpressure gauge.


57

Set the pressure regulator PR2 to obtain 420 kPa (60 psi) on the pressuregauge PG2.

23. Place the system in the closed-loop mode. To do so, connect a leadbetween the 0-10 V output of the [0-24 V] E/E converter and the negat-ive input of the ERROR DETECTOR.

24. Measure the voltage at the ERROR DETECTOR output. Is the voltage null,indicating that the cylinder rod position corresponds to the position calledby the SETPOINT potentiometer 2? Explain.

25. Simulate an increase in cylinder load by setting the Pressure RegulatorPR2 to obtain 560 kPa (80 psi) on the Pressure Gauge PG2. Is the errorautomatically eliminated at the ERROR DETECTOR output?

Yes No

26. Simulate a decrease in cylinder load by setting the Pressure RegulatorPR2 to obtain 280 kPa (40 psi) on the Pressure Gauge PG2. Is the errorvoltage at the ERROR DETECTOR output still null?

Yes No

27. Do your observations confirm that the proportional-plus-integral control isappropriate for eliminating the error in systems where the load varies?

Yes No

Effect of an increase in the integral gain on the step response of the rod position

28. Connect the DC voltmeter at the feedback ( ) input of the ERRORDETECTOR.

29. Set the INTEGRAL GAIN to MIN. and select SETPOINT potentiometer 1.

With a stopwatch in hand, select SETPOINT 2 and measure the time ittakes for the feedback voltage to stabilize to the 5-V setpoint. Record your


58

result in the appropriate row of the column FEEDBACK VOLTAGESTABILIZATION TIME in Table 5-1.

Also, record whether or not the feedback voltage overshoots (exceeds) thesetpoint before it stabilizes.

INTEGRAL GAINPOTENTIOMETER SETTING

FEEDBACK VOLTAGESTABILIZATION TIME

OVERSHOOT THE 5-V SETPOINTBEFORE STABILIZATION?

MIN.

½ of MAX.

¾ of MAX.

MAX.

Table 5-1. Response of the feedback voltage for different INTEGRAL GAIN settings.

30. Repeat the previous step for the other INTEGRAL GAIN settings listed inTable 5-1 and record your results.

31. Based on the data recorded in Table 5-1, what effect does increasing theintegral gain have on the stabilization time and overshooting of thefeedback voltage? Explain.

32. Now experiment further with proportional-plus-integral control.

Set the INTEGRAL GAIN at ½ of MAX. Vary the PROPORTIONAL GAINsetting from MIN. to MAX. by increments of ¼ of a turn.

For each setting, measure the time it takes for the feedback voltage tostabilize to the 5-V setpoint. Also, observe if the feedback voltageovershoots the setpoint before it stabilizes.

33. Based on your observations, what effect does increasing the proportionalgain have on the stabilization time and overshooting of the feedbackvoltage? Explain.


59

34. Now experiment further with integral control. Remove the proportionalamplifier from the circuit by removing the two leads which are connectedat its terminals.

Connect the ERROR DETECTOR output to the integral amplifier input, andset the INTEGRAL GAIN at ½ of MAX.

Alternately select SETPOINT potentiometers 1 and 2 and observe theresponse of the cylinder rod before it stabilizes to the position called by the5-V setpoint.

35. Do your observations confirm the theoretical response shown in Fig-ure 5-7?

Yes No

36. Measure the voltage at the ERROR DETECTOR output. Is the voltage null,indicating that the cylinder rod position corresponds to the position calledby the SETPOINT potentiometer 2?

Yes No

37. On the Conditioning Unit, close the shutoff valves, and turn the regulatoradjusting knob completely counterclockwise.

38. Turn off the PID Controller and the DC Power Supply.


CONCLUSION

In this exercise, you studied proportional-plus-integral control of cylinder rodposition. You saw that unlike proportional control, proportional-plus-integral controlautomatically eliminates the error between the setpoint and feedback voltage(measured position) due to the integral action of the controller.

You observed the effect of increasing the integral gain and proportional gain on theresponse of the rod position (feedback voltage) to a sudden change in setpoint.You saw that as the integral or proportional gain is increased, the stabilization timedecreases. At excessive integral or proportional gains, the system may even startto oscillate without being able to return to equilibrium.


60

REVIEW QUESTIONS

1. What does “integral gain” mean?

2. What are the advantages and disadvantages of integral control?

3. Briefly describe the operation of a controller in the proportional-plus-integralmode.

4. What effect does increasing the integral gain have on proportional-plus-integralposition control?

5. What is the purpose of the anti-reset function?

Other Samples Extracted

from


63

Unit Test

1. When calculating the total piston area for the rod end of a cylinder

a. the rod area must be added to the piston area.b. the rod area must be subtracted from the piston area.c. the piston area must be subtracted from the rod area.d. the piston area must be added to the rod area.

2. A directional control valve with two flow path configurations and three fluidports is called

a. 3-way, 2-position directional control valve.b. 2-way, 3-position directional control valve.c. 2-way, 2-position directional control valve.d. 3-way, 3-position directional control valve.

3. If the same pressure is applied to both sides of the piston in a double-actingcylinder, the cylinder will

a. extend because the extension force is greater than the retraction force.b. extend because the seals in the cylinder will allow fluid in the rod end to

leak into the cap end.c. retract because the pressure applied to the cylinder rod will create a

retraction force greater than the extension force.d. remain in its original position because pressure in each side will prevent

the piston from moving.

4. The devices that convert fluid energy into mechanical energy are called

a. air bearings.b. pneumatic motors.c. cylinders.d. actuators.

5. They are commonly used to control the speed of fluid actuators.

a. Directional control valves.b. Check valves.c. Flow control valves.d. Shutoff valves.

6. The symbol for a directional control valve consists of a separate envelope foreach

a. valve.b. flow path.c. way.d. position.

Unit Test (cont'd)

64

7. They are commonly used to control the direction of fluid actuators.

a. Check valves.b. Directional control valves.c. Flow control valves.d. Shutoff valves.

8. When the actuator does not constantly work against a load,

a. the meter-in configuration is commonly used to control the speed.b. the meter-in and meter-out configuration is commonly used to control the

speed.c. the meter-out configuration is commonly used to control the speed.d. the by-pass configuration is commonly used to control the speed.

9. When an intensifier is contained within an individual cylinder,

a. the rod end is used as the low-pressure side.b. the rod end is used as the high-pressure side.c. the cap end is used as the high-pressure side.d. the rod end is used as the master cylinder.

10. If two cylinders connected in series are of the same size and stroke, theupstream cylinder will extend

a. faster.b. slower.c. at exactly the same speed.d. after the downstream cylinder.

Instructor’s Guide Sample

Extract from

Electrical Control

of Pneumatic Systems

Industrial Applications

67

EXERCISE 4-1 PNEUMATIC ACTUATOR DECELERATIONCIRCUITS

ANSWERS TO PROCEDURE QUESTIONS

10. When the rod is in the fully retracted position, LS1 is actuated. NO contactLS1 and NC contact LS2 are closed. Pressing PB1 causes DV1-SOL-A tobe energized and the cylinder rod extends.

11. No. Flow control valve FCV1 is fully open and does not restrict air flow todecelerate the cylinder rod.

14. Yes. The air forced out of the cylinder rod end is restricted by FCV1 andthe extension speed is slowed down.

15. Because of the bypass check valve of the flow control valve FCV1.

16. The cylinder rod will not complete a full extension and the system will stopif the rod does not reach LS2.

25. When the START pushbutton PB1 is pressed, relay coil CR1 is energized.This causes NO contact CR1-B to go closed, DV1-SOL-A to be energizedand the motor to start rotating.

26. No. The air flow is block by check valve CV1 in one direction.

27. When the BRAKE pushbutton PB3 is pressed, relay coil CR2 is energized.This causes CR1 to be de-energized. This also causes the spool of DV1to shift and the motor to stop.

29. When the STOP pushbutton PB2 is pressed, relay coil CR1 is de-energized and the spool of DV1 returns to its center position. This causesthe motor to stop rotating. The motor acts like a vacuum pump whiledecelerating.


68

ANSWERS TO REVIEW QUESTIONS

1. The check valve is used to allow the retraction of the cylinder rod at full speed.

2. The degree of cushioning can be adjusted with the needle valve.

3. The cylinder rod extends at full speed until the sensing device detects thepresence of the rod. At this point, the air forced-out of the cylinder rod end isredirected through a flow control valve that restricts the flow.

4. A bi-directional pneumatic motor may be stopped without pulling a vacuum,using a particular center configuration of the directional valve as shown inFigure 4-5.

5. Because of the compressibility of air.

EXERCISE 4-2 COUNTING OF ACTUATOR CYCLES


4. The display count should be 000.

5. The displayed count is incremented by one. Pressing PB1 momentarilyswitches the CONTROL input of the counter to common. This activates theCONTROL input and increments the counter value by one.

6. Pilot lamp L1 should turn on. When the counter reaches the preset value,the counter output contacts are shifted to their activate state. This causesCT1-A to close, energizing L1.

7. Yes.

8. The displayed count should return to zero and the pilot lamp L1 should turnoff. Pressing the reset button momentarily switches the RESET input of thecounter to common. This resets the count to zero and deactivates CT1-A,de-energizing L1.

9. The display count is incremented after PB1 is released.

10. Yes.


69

19. When the START pushbutton PB1 is pressed, the current is allowed to flowto energize relay coil CR1 and SOL-A of directional valve DV1. Thiscauses the spool of DV1 to shift and the cylinder rod to extend.

20. Energizing relay coil CR1 causes NO contact CR1-C on rung 3 to goclosed. The COMMON input of the counter is then connected to commonand the count value increases by one.

21. Yes. This is because when the cylinder completes its fifth extension stroke,the count value of counter CT1 reaches the preset value of 5, whichactivates the counter relay contacts. This causes NO contact CT1-B onrung 3 to close, connecting the counter RESET (R) input to common andcausing counter CT1 to reset its count to zero.

22. Yes.

23. Yes. Pressing the STOP pushbutton PB3 causes relay coil CR2 to de-energize. This causes holding contact CR2-A to go open. When thecylinder is created on rung 1 and the cylinder stops.

24. The cylinder rod does not extend. The START pushbutton PB1 creates anopen-circuit condition on rung 1. This prevents DV1-SOL-A and CR1 frombeing energized to start the cylinder extension.

29. Yes.

30. Yes.

40. Motor speed: 2000 ±500 r/min

41. Yes.


1. Electrical counters are used to count quantities produced during process andcontrol operation. They are also used for machine maintenance scheduling bycontrolling the number of machine operations.

2. Each time a pulse is received at the control terminal, the counter count isincremented by one. When the preset count is reached, the counter activatesits output contacts. Additional inputs continue to increment the count.


70

Momentary activation of the reset input deactivates the counter contacts andresets the count to zero.

3. The preset value is the count the counter must reach before activating itsoutput contacts.

4. Yes.

5. The counter value can be returned to zero by momentarily activating the resetinput of the counter.

EXERCISE 4-3 INDUSTRIAL DRILLING SYSTEM AND SAFETYCIRCUITS


8. Yes.

10. When the START pushbutton PB1 is pressed, the current is allowed to flowthrough ladder rung 1 to energize relay coil CR1. DV1-SOL-A is energized.This shifts the valve causing the clamp cylinder to extend and thepneumatic drill unit to rotate.

11. When the clamp cylinder becomes fully extended, the pressure risesquickly behind its piston. When it reaches the actuation 600 kPa (80 psi),it activates pressure switch PS1, causing NO contact PS1 on rung 1 to goclosed. This energizes DV2-SOL-A and shifts the spool of the valve andthe drill cylinder extends.

12. When the drill cylinder becomes fully extended, it activates photoelectricswitch PE1. This causes NC contact PE1 on rung 1 to go open, de-energizing relay coil CR1. This causes DV2-SOL-A to de-energize and thedrill cylinder retracts. DV1-SOL-A is also de-energized but the spool ofDV3 does not shift. So the pneumatic motor continues to rotate.

13. When the drill cylinder becomes fully retracted, it activates magneticproximity switch PX2. This causes NC contact PX2 on rung 2 to go open.This de-energizes relay coil CR2 causing DV1-SOL-B to energize and thespool of DV1 to shift. This causes the pneumatic motor to stop and theclamp cylinder to retract at full speed.

20. No.


71

21. No.

22. The cylinder rod extends full stroke.

23. Yes.

24. The cylinder rod immediately retracts.

25. Yes. You would be able to operate the cylinder circuit simply by pressingthe other pushbutton, PB2

31. Yes.

32. The cylinder rod does not extend. Pressing pushbutton PB1 causes relaycoil CR1 to energize, opening contact CR1-B and closing contact CR1-A.Since contact CR2-B is closed, this also causes timer TD1 to start timing.1 second later, relay contacts TD1-A and TD1-B open, de-energizing relaycoil CR1 and preventing relay contacts CR1-A and CR1-B from closing toenergize valve solenoid SOL-A.

33. If PB1 is pressed and then PB2 is pressed, the time which can elapsebetween the two events before the cylinder fails to operate is the time seton the time-delay relay. In this case, 1 s.

34. No. Tying down PB2 would cause timer TD1 to energize throughNO contact PB2 and NC contact CR1-B in the closed condition. 1 secondlater, TD1 would shift its NC contacts TD1-A and TD1-B to the opencondition. This would prevent relay coils CR1 and CR2 from beingenergized and keep relay contacts CR1-A and CR2-A in the opencondition until both pushbuttons are released.


1. The drill cylinder will not extend after extension of the clamp cylinder. Thepressure developing at the pneumatic port of PS1 will be insufficient to activatePS1.

2. The pushbuttons have to be located far enough apart so that two hands arerequired for operating them.


72

3. No, because the operator would be unable to instantly retract the cylinderduring its extension if a hazardous situation should occur.

4. A two-hand, non-tie-down safety circuit is a circuit that requires the use of bothhands to start the circuit and that cannot be override by tying down one of thepushbuttons.

5. The cylinder rod does not extend.


73

(+)

PB1

(−)

L1

1 CR1

DV1-SOL-A

CR1-A

DV1-SOL-B

3

START

L2

DV2-SOL-A

CR2

2

TD1, NO

PS1, NO

PX1, NO

LS2, NO

CR2-A

TD1+R

C

−

LS1, NO

LS1, NC

CR1-B

L3

CR2-B

EXERCISE 4-4 GARBAGE COMPACTOR SIMULATIONCIRCUIT


1.

Figure 4-20.

9. Yes.

10. No. Depending on the garbage volume, the length of extension of thedepress cylinder will vary.


74

11. The garbage will be less compacted.

12. To maintain pressure on the garbage a preset time to improve compaction.


1. Magnetic proximity switch PX1 is used to confirm the retracted position of thedepress cylinder.

2. The garbage would be more compacted.

3. A pressure regulator.

4. The circuit will not start. Limit switch LS1 must confirm the retracted position ofthe feed cylinder to operate the system.

5. LS2.

ANSWERS TO UNIT TEST

1. b; 2. c; 3. a; 4. d; 5. b; 6. d; 7. a; 8. d; 9. c; 10. c.

Bibliography

Hedges, Charles S., Industrial Fluid Power, Volume 1, Third Edition, Dallas, Texas:Womack Educational Publications, Department of Womack Machine SupplyCompany, 1993

ISBN 0-9605644-5-4

Hedges, Charles S., Industrial Fluid Power, Volume 2, Fourth Edition, Dallas,Texas: Womack Educational Publications, Department of Womack Machine SupplyCompany, 1988

ISBN 0-943719-01-1

Parker Hannifin Corporation. Industrial Pneumatic Technology. Bulletin 0275-B1

Documents

Pneumatics Training System, Model 6081